Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{2014}\)

\(\Rightarrow-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(-A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

\(\Rightarrow-2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

\(\Rightarrow-2A-\left(-A\right)=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\right)\)

\(-A=2-\frac{1}{2^{10}}\)

\(\Rightarrow A=\frac{1}{2^{10}}-2\)

Hửhh

2A=1+1/2+1/4+1/8+.........+1/512

2A‐A=﴾1+1/2+1/4+1/8+....+1/512﴿‐﴾1/2+1/4+1/8+.....+1/1024﴿

A=1‐1/1024 =1023/1024

vậy A=1023/1024

Đặt A=\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.........+\frac{1}{1024}\) (1)

Ta có:  2A=\(2+1+\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{512}\) (2)

Từ (1) và (2) \(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...........+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+..........+\frac{1}{1024}\right)\)

\(\Rightarrow A=2-\frac{1}{1024}\)

\(\Rightarrow A=\frac{2047}{1024}\)

\(\text{Đặt }A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(=\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

=> \(\frac{1}{2}A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\)

=> \(\frac{1}{2}A-A=\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

=> \(-\frac{1}{2}A=\frac{1}{2^{11}}-1=\frac{1-2048}{2048}=-\frac{2047}{2048}\)

=> \(A=-\frac{2047}{2048}:\left(-\frac{1}{2}\right)=\frac{2047}{1024}\)

1,2 : 0,5 - 1/4 : 0,25 + 1/8 ; 0,125 - 1/10 : 0,1

= 1/2 x 1/0,5 - 1/4 x 1/0,25 + 1/8 x 1/0,125 - 1/10 x 1/01

=       1 - 1 + 1 - 1

= 0

anh_hung_lang_la thì làm đi

bọn kia có trả lời câu hỏi không thì bảo

1)=1023/1024

2)=1093/729

Ta có: \(-1=-2+1;-\frac{1}{2}=-1+\frac{1}{2};-\frac{1}{4}=-\frac{1}{2}+\frac{1}{4};...;-\frac{1}{1024}=-\frac{1}{512}+\frac{1}{1024}\)

\(\Rightarrow-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(=\left(-2+1\right)+\left(-1+\frac{1}{2}\right)+\left(-\frac{1}{2}+\frac{1}{4}\right)\)\(+...+\left(-\frac{1}{512}+\frac{1}{1024}\right)\)

\(=-2+1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-...-\frac{1}{512}+\frac{1}{1024}\)

\(=-2+\frac{1}{1024}\)

\(=-\frac{2047}{1024}\)

Đặt: \(A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)

\(\Rightarrow\dfrac{4}{2}A=\dfrac{4}{2}\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\right)\)

\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\)

\(\Rightarrow2A-A=\left(3+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)

\(\Rightarrow A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(3-1-\dfrac{1}{1024}\right)\)

\(\Rightarrow A=2-\dfrac{1}{1024}\)

\(\Rightarrow A=\dfrac{2047}{1024}\)

 1+(1)/(2)+(1)/(4)+(1)/(8)+...+(1)/(512)+(1)/(1024)

A x 2 = 1 - ( 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/512 + 1/1024 ) - 1/1024

A x 2 = 1 - 1/1024 + A

A x 2 - A = 1 - 1/1024

A = 1 - 1/1024 

A = 1023 /1024

-1-1/2-1/4-1/8......-1/1024

=-(1+1/2+1/4+1/8...+1/1024)

mà ta có 1024=2^10

nên -(1+1/2+1/4+1/8...+1/1024)

=-(2^9+2^8+2^7....+1)/2^10

=-(1023/1024)

=-1,99.........

mình sẽ làm lại bai này cho đúng nha

\(-1-\frac{1}{2}-\frac{1}{4}....-\frac{1}{1024}=-1-\left(\frac{1}{2}+\frac{1}{4}+...\frac{1}{1024}\right)\)

\(=-1-\left(\frac{1}{2^1}+\frac{1}{2^2}...+\frac{1}{2^{10}}\right)\)

\(=-1-\frac{1023}{1024}=\frac{-1024}{1024}-\frac{1023}{1024}=\frac{-2047}{1024}\)

vậy mới đúng nha

ta có\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

tách

\(B=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(2B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)

\(2B-B=\frac{1}{2}-\frac{1}{1024}\)

thay vào B ta có 

\(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)

\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\)

\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)

\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2^{10}}\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^{10}}\)

\(\Rightarrow A=\frac{2^9+1}{2^{10}}\)

\(\Rightarrow A=\frac{513}{1024}\)